mult value calculation for the DFT-NEB calculation
Just Got Here
1:01:04 PM PST - Wed, Dec 4th 2013
Hi NWChem users,

I am trying to do DFT-NEB calculation of C-N reaction. It needs ‘mult’ keyword since there are odd number of electrons in the system. ‘mult’ is the spin multiplicity of the system. I found there are two
ways to calculate the 'mult' values.
One is ‘number of alpha electrons (i.e., electron with the spin quantum number of 1/2) minus beta electrons (i.e., electron with the spin quantum number of -1/2) plus 1’ other is ‘2S+1’ where S is the 0.5*(number of unpaired electrons).
Using the both definition, I got mult = 6 for the C-N system. I am wondering whether I am doing any mistake. Here is the DFT-NEB input script.

start neb-c-n
basis 
 * library 6-311++G**
end

dft
  xc b3lyp

  maxiter 5000

  cgmin

  mult 6
end

        1. define the start geometry ####
geometry units angstroms print xyz noautosym
C 39.59843 39.09966 39.40598
N 39.26613 41.94704 40.23538
end

        1. define the end geometry ####
geometry endgeom units angstroms print xyz noautosym
C 39.20759 39.88065 39.43266
N 39.08069 41.28918 39.77152
end

neb
  maxiter 1000

  nbeads 20

  stepsize 0.10

  kbeads 0.1
end
task dft neb ignore

I will appreciate your comments regarding the mult value calculation and input script.

Thank you.
kind regards,
Sanjib

Forum Regular
3:17:34 PM PST - Wed, Dec 4th 2013
Hi Sanjib,

This is a doublet. There's just one unpaired electron and hence the multiplicity is 2*1/2 + 1 = 2
mult = 2

Best,
-Niri

Just Got Here
8:34:54 AM PST - Thu, Dec 5th 2013
Hi Niri,

Many thanks for your response. I didn’t understand how the number of unpaired electron becomes one in this case. Electronic structure of C and N are 1s22s22px12py1 and 1s22s22px12py12pz1. So in the C atom there are 2 unpaired electrons in the 2px and 2py orbitals whereas in the N atoms there are 3 unpaired electrons in the 2px, 2py and 2pz orbital. Therefore in my sense, total unpaired electron is 2+3 = 5.

Is it like that there are total 1 (in 2px orbital of the C) + 1 (in 2px orbital of the N) = 2 electron in the 2px orbital of the C and N system and so they are paired?

Thank you.
Kind regards,
Sanjib

Just Got Here
5:01:58 PM PDT - Mon, Mar 10th 2014
Hi Sanjib,

Your occupation numbers are correct for carbon and nitrogen atoms, but things are different when you switch to molecular orbitals in the CN radical. The simplest explanation is probably through orbital hybridization. A quick internet search or chemistry textbook would give more details, but in short:

You can ignore the core electrons (the 1s electrons) from each atom and focus on the valence electrons. There are 4 valence electrons in carbon and 5 valence electrons for nitrogen. Since there are only two atoms, the molecule will be linear with sp hybridized orbitals mixing contributions from the 2s orbitals and one of the 2p orbitals on each atom. You will have two sp hybrid orbitals and two remaining 2p orbitals. One pair of sp orbitals (one from each atom) will overlap to form a sigma bond with one electron from each atom. Both pairs of 2p orbitals will overlap to form two pi bonds (with each bond having one electron from each of the atoms). The sp orbital left on nitrogen will be doubly occupied (lone pair) because nitrogen will be surrounded by 5 valence electrons (one from each bond and both from the lone pair) to maintain a neutral charge. The final electron will be unpaired in the remaining sp orbital on carbon.

The final result is a triple bond between the two atoms, a lone pair on the nitrogen, and a single unpaired electron on carbon. 2S+1 = 2

Cheers,
Matt


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